In my notation, $\langle\bullet,\bullet\rangle$ is linear in the first variable and anti-linear in the second variable. Let $Q:H\to H$ be defined by
$$ Qh=h-\langle h,e_1\rangle e_1-\langle h,e_2\rangle e_2$$ for all $h\in H$. Note that $Q$ is hermitian because
\begin{align}\langle Qu,v\rangle &=\big\langle u-\langle u,e_1\rangle e_1-\langle u,e_2\rangle e_2,v\big\rangle\\&=\langle u,v\rangle -\langle u,e_1\rangle \langle e_1,v\rangle -\langle u,e_2\rangle \langle e_2,v\rangle\\&=\langle u,v\rangle -\langle u,e_1\rangle \overline{\langle v,e_1\rangle} -\langle u,e_2\rangle \overline{\langle v,e_2\rangle}\\&=\big\langle u,v-\langle v,e_1\rangle e_1-\langle v,e_2\rangle e_2=\langle u,Qv\rangle.\end{align}
Next, we prove that $Q$ is a projection. That is, $Q^2=Q$. To show this, let $h\in H$ be arbitrary. We have
\begin{align} Q^2h&=Q(Qh)=Q\big(h-\langle h,e_1\rangle e_1-\langle h,e_2\rangle e_2\big)\\&=Qh-\langle h,e_1\rangle Qe_1-\langle h,e_2\rangle Qe_2\\&=Qh-\langle h,e_1\rangle \big(e_1-\langle e_1,e_1\rangle e_1-\langle e_1,e_2\rangle e_2\big) -\langle h,e_2\rangle \big(e_2-\langle e_2,e_1\rangle e_1-\langle e_2,e_2\rangle e_2\big) \\&=Qh-\langle h,e_1\rangle (e_1-e_1-0)-\langle h,e_2\rangle (e_2-0-e_2\rangle\\&=Qh-\langle h,e_1\rangle \cdot 0-\langle h,e_2\rangle\cdot 0= Qh.\end{align}
Now, observe that $Qe_k=e_k$ for $k=3,4,5,\ldots$ but $Qe_1=Qe_2=0$. Therefore, for any $h\in H$, $Qh\perp e_1$ and $Qh\perp e_2$. This is because
$$\langle Qh,e_k\rangle =\langle h,Qe_k\rangle =\langle h,0\rangle =0$$
for $k=1,2$, so $Qh\in \{e_1,e_2\}^\perp =E$. This proves that $\operatorname{im}Q\subseteq E$. The final task to show that for any $h\in E$, $Qh=h$, and this establishes the claim that $\operatorname{im} Q=E$. That is, $Q=P_E$. To see this, we suppose that $h\in E$. Thus, $h\perp e_1$ and $h\perp e_2$, so $\langle h,e_1\rangle=\langle h,e_2\rangle=0$. That is,
$$Qh=h-\langle h,e_1\rangle e_1-\langle h,e_2\rangle e_2=h-0e_1-0e_2=h.$$
I think it is generally true that if $\{e_1,e_2,e_3,\ldots\}$ is an orthonormal basis of a separable Hilbert space $H$ and $P$ is the orthogonal projection onto a closure of the subspace spanned by $\{e_k:k\in A\}$, where $A$ is a subset of $\Bbb N_1$, then $$Ph=\sum_{k\in A}\langle h,e_k\rangle e_k=h-\sum_{k\in \Bbb N_1\setminus A} \langle h,e_k\rangle e_k$$
for all $h\in H$. In other words, $P$ is the projection onto the orthogonal complement of $\{e_k:k\in\Bbb{N}_1\setminus A\}$.
Say the family of orthonormal vectors is indexed by some set $I$. For any finite $F \subseteq I$ we have $$0 \leq \bigg\lVert f - \sum_{i \in F} \langle f, e_i \rangle e_i \bigg\rVert^2 = \lVert f \rVert^2 - 2\sum_{i \in F} |\langle f, e_i \rangle|^2 + \sum_{i \in F} |\langle f, e_i \rangle|^2 = \lVert f \rVert^2 - \sum_{i \in F} |\langle f, e_i \rangle|^2.$$
So, $\sum_{i \in F} |\langle f, e_i \rangle|^2 \leq \lVert f \rVert^2$ for all finite $F \subseteq I$. Thus $\sum_{i \in I} |\langle f, e_i \rangle|^2 \leq \lVert f \rVert^2 < \infty$. In particular, the set
$$\{i \in I : \langle f, e_i \rangle \neq 0\} = \bigcup_{n=1}^\infty \{i \in I : |\langle f, e_i \rangle| > 1/n\}$$
is countable since for each $n$ the set $\{i \in I : |\langle f, e_i \rangle | > 1/n\}$ is finite. Let $(e_j)_{j=1}^\infty$ be an enumeration of those $e_i$'s for which $\langle f, e_i \rangle \neq 0$. Then for $m \leq k < \infty$,
$$\bigg\lVert \sum_{j=m}^k \langle f, e_i \rangle e_j \bigg\rVert^2 = \sum_{j=m}^k |\langle f, e_j \rangle|^2 \to 0 \text{ as } k,m \to \infty.$$
In other words, the partial sums of $\sum_{j=1}^\infty \langle f,e_j \rangle e_j$ are Cauchy, so the series converges by completeness of $H$.
Best Answer
Note that $$ V = \left\{ f \in L^2 : \int_0^1 fdx = 0 \right\} = [\{ 1\}]^{\perp}. $$ So $L^2=V\oplus [\{1\}]$. The orthogonal projection of $f$ onto $[\{1\}]$ is $\langle f,1\rangle 1$, and the orthogonal projection of $f$ onto $V$ is $f-\langle f,1\rangle 1$.